Show that sum of the two closed sets in $\mathbb{R^2}$ is not always closed.

In $\mathbb{R^2}$, consider the sets $A =\{(x,y)\,\,:\,\, x>0,\,\,xy=1\}$ and $B =\{(x,y)\,\,:\,\, x>0,\,\,xy=-1\}$ which are closed, show that the sum A + B is not closed. $$ A+B:=\{a+b : a \in A, b \in B\} $$

To have better intuition, you can visualize the two sets as the following picture

enter image description here

  • 3
    $\begingroup$ ${}$Nice example! $\endgroup$ – Lord Shark the Unknown Oct 10 '18 at 18:45
  • 1
    $\begingroup$ What's the question? $\endgroup$ – Yanko Oct 10 '18 at 18:56
  • $\begingroup$ I revised the question. Show that the set $A+B$ is not closed. $\endgroup$ – Saeed Oct 10 '18 at 19:03
  • $\begingroup$ What subset of the plane is $A+B$? $\endgroup$ – rogerl Oct 10 '18 at 19:56
  • $\begingroup$ What do you mean? $\endgroup$ – Saeed Oct 10 '18 at 23:11

Consider the sequence $$(x_n)=\Big\{\Big(\frac{2}{n},0\Big)\Big\}_{1}^\infty=\Big\{\Big(\frac{1}{n},n\Big)+\Big(\frac{1}{n},-n\Big)\Big\}_{1}^\infty \in A+B$$ Here $$x_n \longrightarrow (0,0) \notin A+B$$ since the first coordinate of the element in $A+B$ is always positive .

So, $A+B$ is not closed!


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